Question

Laplace transform of \( g(t) = \begin{cases} \cos(t-\frac{\pi}{3}), & \text{if } t>\frac{\pi}{3} \\ 0, & \text{if } t<\frac{\pi}{3} \end{cases} \) is

• A. \( \frac{s e^{-s\pi/3}}{s^2+1} \)
• B. \( \frac{e^{-s\pi/3}}{s^2-1} \)
• C. \( \frac{- \pi s e^{\frac{s\pi}{3}}}{s^2+1} \)
• D. \( \frac{\pi s e^{\frac{s\pi}{3}}}{s^2-1} \)

Hint:

Laplace Transform - Second Shifting Theorem. If \(\mathcal{L\{f(t)\ = F(s)\), then \(\mathcal{L\{f(t-a)u(t-a)\ = e^{-asF(s)\). Also, \(\mathcal{L\{\cos(\omega t)\ = s/(s^2+\omega^2)\).

Answer 1

The Correct Option is A

The function \(g(t)\) can be written using the Heaviside unit step function \(u(t-a)\): $$ g(t) = \cos\left(t - \frac{\pi}{3}\right) u\left(t - \frac{\pi}{3}\right) $$ This is of the form \( f(t-a) u(t-a) \), where \( f(t) = \cos(t) \) and \( a = \pi/3 \).
We use the Second Shifting Theorem (Time Delay Theorem) of Laplace transforms: $$ \mathcal{L}\{ f(t-a) u(t-a) \} = e^{-as} F(s) $$ where \( F(s) = \mathcal{L}\{ f(t) \} \).
First, find the Laplace transform of \( f(t) = \cos(t) \).
The standard transform is: $$ F(s) = \mathcal{L}\{\cos(t)\} = \frac{s}{s^2 + 1^2} = \frac{s}{s^2 + 1} $$ Now apply the Second Shifting Theorem with \( a = \pi/3 \): $$ G(s) = \mathcal{L}\{ g(t) \} = e^{-(\pi/3)s} F(s) = e^{-s\pi/3} \frac{s}{s^2 + 1} $$ $$ G(s) = \frac{s e^{-s\pi/3}}{s^2+1} $$ This matches option (1).